A McBurger meal includes a choice of two different side dishes out of 5 and a selection of one drink from 4 options. How many different meals could a patron select?

The side choices: You have 5 options, and you have to choose 2. For your first side dish, you have 5 choices. After you choose your first dish, you can only choose your second dish among the 4 leftover choices (because you don't want your two dishes to be the same!). So first dish you have 5 choices, second dish you have 4 choices, together you have 5 * 4 = 20 options for dishes right? Not quite. See, if you choose option A and then choose option B, it's the same as choosing option B and then option A, you end up with the same meal! In fact, Every combination in the 20 combinations we just calculated has one doppelgänger. That's easy, we just divide the number 20 by 2 to get 10 unique combinations. I'll list them here: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.

Now, you have 10 choices for your meal and 4 choices for your drink, so together you have 10 * 4 = 40 combinations. Can quite fit the whole list here, but imagine the list above with meal options, and each of those options have 4 drink options, like: AB(a) AB(b) AB(c) AB(d) AC(a) AC(b) AC(c) AC(d) AD(a).....

Hope I helped!